Question

s(t) = 16t^3 + 12t^2-144t. t>0. On what open interval(s) is the particle traveling in the negative direction?

Answer 1

So we need to know for what values of t is v(t)<0
\[ s(t) = 16t^3 + 12t^2-144t\]\[v(t)=s'(t) = 48t^2 + 24t-144=0\]when the particle is changing direction, so solving for t gives\[2t^2+t-6=0=(2t-3)(t+2) \rightarrow t=\left\{ -2,3/2 \right\}\]since we are given that t>0 the only solution to worry about is t=3/2 (this is when the particle changes direction.

So the particle is going one direction when
0<t<3/2
and another afterward, i.e. when
t>3/2
so plug in some value of t in the into v(t) that is in either interval and see if v(t) is positive or negative. Let's use t=1:\[v(1)=48+24-144=-72<0\]and since t-=1 is in the interval

0<t<3/2

that means that is when the particle is moving in the negative direction.

Answer 2

In interval notation\[(0,3/2)\]is when the particle is moving in the negative direction.